2.
Factorise each of the following expressions
completely
(a) 3n2 + 10n + 7
(b) 4p2 + 8p + 3
(c) 6q2 - 179 + 12 (d) 4r2 - 7r + 3
(e) 852 + 2s - 15
(f) 6t2 + 19t - 20
(g) 4u2 - 8u - 21
(h) 18w2 - W - 39
Answers
Answer:mark me brainliest
Step-by-step explanation:
A)Step-1 : Multiply the coefficient of the first term by the constant 3 • 7 = 21
Step-2 : Find two factors of 21 whose sum equals the coefficient of the middle term, which is -10 .
-21 + -1 = -22
-7 + -3 = -10 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -7 and -3
3n2 - 7n - 3n - 7
Step-4 : Add up the first 2 terms, pulling out like factors :
n • (3n-7)
Add up the last 2 terms, pulling out common factors :
1 • (3n-7)
Step-5 : Add up the four terms of step 4 :
(n-1) • (3n-7)
Which is the desired factorization
Final result :(n-1) * (3n-7)
using this peocess do all the problems.if i do all it will take lots of time.sry
Answer:
Step-by-step explanation:
(a) 3n² + 10n + 7
3n² + 3n + 7n + 7
3n(n + 1) + 7(n + 7)
(3n + 7)(n + 7)
3n + 7 = 0
n = -7/3
n + 7 = 0
n = -7
(b) 4p² + 8p + 3
4p² + 2p + 6p + 3
2p(2p + 1) + 3(2p + 1)
(2p + 3)(2p + 1)
2p + 3 = 0
p = -3/2
2p + 1 = 0
p = -1/2
(d) 4r² - 7r + 3
4r² - 4r - 3r + 3
4r(r - 1) - 3(r - 1)
(4r - 3)(r - 1)
4r - 3 = 0
r = 3/4
r - 1 = 0
r = 1
(f) 6t² + 19t - 20
6t² + 24t - 5t - 20
6t(t + 4) - 5(t + 4)
(6t - 5)(t + 4)
6t - 5 = 0
t = 5/6
t + 4 = 0
t = -4
(g) 4u² - 8u - 21
4u² - 14u + 6u - 21
2u(2u - 7) + 3(u - 7)
(2u + 3)(2u - 7)
2u + 3 = 0
u = -3/2
2u - 7 = 0
u = 7/2
(h) 18w² - w - 39
18w² + 26w - 27w - 39
2w(9w + 13) - 3(9w + 13)
(2w - 3)(9w + 13)
2w -3 = 0
w = 3/2
9w + 13 = 0
w = -13/9
I hope this answer is helpful for you. I can't solve (c) and (e) part because your question is not clear.²